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16 COMPACT SPACES 57

from (3.15.2)). Conversely, suppose the condition satisfied, and let us prove
that the mapping/ defined by J(x) = lim f(y) is a solution of the extension

y-*jc,y e A

problem. First of all, if x e A, the existence of the limit implies by definition
/(*) =/(X)> ^hence / extends /, and it remains to see that / is continuous.
Let x e E, V a neighborhood of f(x) in E'; there is a closed ball B' of center
f(x) contained in V. By assumption, there is an open neighborhood V of x
in E such that/(V n A) e B' (by (3.13.1)). For any y e Vj(y) is the limit of
/at the point y with respect to A, hence also with respect to V n A, by (3.13.5);
hence, it follows from (3.13.7) that/00 e/(V n A), and therefore f(y) e B'
since B' is closed. Q.E.D.

(3.15.6) Let A be a dense subset of a metric space E, and f a uniformly
continuous mapping of A into a complete metric space E'. Then there exists
a continuous mapping J of E into E' coinciding with f in A; moreover, f is
uniformly continuous.

To prove the existence of/, it follows from (3.15.5) and (3.14.6) that we
have to show the oscillation of / at any point x e E, with respect to A,
is 0. Now, for any 8 > 0, there is 6 > 0 such that d(y, z) < 8 implies
d'(f(y\f(z)) < e/3 (y, z in A). Hence, the diameter of/(A n B(;c; 5/2)) is at
most e/3, which proves our assertion. Consider now any two points s, t in E
such that d(s₉1) < 6/2. There is a y 6 A such that d(s, y) < 6/4 and
d'(f(s),f(y)) < e/3, and a z e A such that d(t, z) < <5/4 and d'(J(t)₉f(z)) < e/3.
From the triangle inequality it follows that d(y, z) < 6, and as y, z are in A,
d'(f(y),f(z)) <e/3; hence, by the triangle inequality, d'(f(s)J(t))<e; this
proves that / is uniformly continuous.

PROBLEM

Let n -*• r_n be a bijection of N onto the set A of all rational numbers x such that 0 < x ^ 1
(2.2.15). We define a function in E « [0, 1] by putting/(x) = £ V2", the infinite sum being

r_n<x

extended only to those n such that r_n < x. Show that the restriction of/to the set B of all
irrational numbers x e [0, 1) is continuous, but cannot be extended to a function continuous
inE.

16. COMPACT SPACES

A metric space E is called compact if it satisfies the following condition
("Borel-Lebesgue axiom")- for every covering (U_A)_AeL of E by open sets



	16 COMPACT SPACES 57 from (3.15.2)). Conversely, suppose the condition satisfied, and let us prove that the mapping/ defined by J(x) = lim f(y) is a solution of the extension y-jc,y e A problem. First of all, if x e A, the existence of the limit implies by definition /() =/(X)> ^hence / extends /, and it remains to see that / is continuous. Let x e E, V a neighborhood of f(x) in E'; there is a closed ball B' of center f(x) contained in V. By assumption, there is an open neighborhood V of x in E such that/(V n A) e B' (by (3.13.1)). For any y e Vj(y) is the limit of /at the point y with respect to A, hence also with respect to V n A, by (3.13.5); hence, it follows from (3.13.7) that/00 e/(V n A), and therefore f(y) e B' since B' is closed. Q.E.D.

	(3.15.6) Let A be a dense subset of a metric space E, and f a uniformly continuous mapping of A into a complete metric space E'. Then there exists a continuous mapping J of E into E' coinciding with f in A; moreover, f is uniformly continuous. To prove the existence of/, it follows from (3.15.5) and (3.14.6) that we have to show the oscillation of / at any point x e E, with respect to A, is 0. Now, for any 8 > 0, there is 6 > 0 such that d(y, z) < 8 implies d'(f(y\f(z)) < e/3 (y, z in A). Hence, the diameter of/(A n B(;c; 5/2)) is at most e/3, which proves our assertion. Consider now any two points s, t in E such that d(s₉1) < 6/2. There is a y 6 A such that d(s, y) < 6/4 and d'(f(s),f(y)) < e/3, and a z e A such that d(t, z) < <5/4 and d'(J(t)₉f(z)) < e/3. From the triangle inequality it follows that d(y, z) < 6, and as y, z are in A, d'(f(y),f(z)) <e/3; hence, by the triangle inequality, d'(f(s)J(t))<e; this proves that / is uniformly continuous.

	PROBLEM Let n -• r_n* be a bijection of N onto the set A of all rational numbers x such that 0 < x ^ 1 (2.2.15). We define a function in E « [0, 1] by putting/(x) = £ V2", the infinite sum being *r_n<x* extended only to those n such that r_n < x. Show that the restriction of/to the set B of all irrational numbers x e [0, 1) is continuous, but cannot be extended to a function continuous inE.

	16. COMPACT SPACES A metric space E is called compact if it satisfies the following condition ("Borel-Lebesgue axiom")- for every covering (U_A)_AeL of E by open sets